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Is there a fast way to find the first offset at which two large byte sequences differ?

I can use a for loop to loop over two byte sequences and return the index at the first difference of course:
bytes1 = b'12345'
bytes2 = b'1F345'
for index, pair in enumerate(zip(bytes1, bytes2)):
if pair[0] != pair[1]:
print(index)
break
But I don't think that's a smart and fast way to do it. I would hope a native method exists that I can call to get this done. Is there something that can help me here? I can also use numpy if it helps.
I also want to clarify that this will run many times, with medium sequences. Approximately 300MB is expected, chunked by 100kB. I might be able to change that for larger if it helps significantly.

a solution with numpy is to convert them to an array of uint8 then xor them and use argmax to get the first non-zero.
import numpy as np
bytes1 = b'12345'
bytes2 = b'1F345'
bytes3 = np.frombuffer(bytes1,dtype=np.uint8)
bytes4 = np.frombuffer(bytes2,dtype=np.uint8)
max_loc = np.flatnonzero(bytes3 ^ bytes4)[0]
print(max_loc)
1
problem is that this still iterates to the end of the string on all functions, it's done in C so it is not too slow, slicing long array into multiple smaller ones can reduce the overhead for very long arrays.
Edit: modified argmax to the correct flatnonzero as pointed by #jasonharper, which throws an indexError if both inputs are equal.

If using numba is ok:
import numba
#numba.jit()
def method2(bytes1, bytes2):
idx = 0
while idx < len(bytes1):
if bytes1[idx] != bytes2[idx]:
return idx
idx += 1
return idx
Note that first run of this function will be significantly slower (due to compilation performed by numba). Takes like 2 seconds.
Then, for each next run of the function:
easy case you posted, index = 1 -> numba is 2x faster,
for index = 100 -> numba is 33x faster

The most efficient way rather than using np.setdiff1d and np.in1d, to remove common values of 1D arrays with unique values

I need a much faster code to remove values of an 1D array (array length ~ 10-15) that are common with another 1D array (array length ~ 1e5-5e5 --> rarely up to 7e5), which are index arrays contain integers. There is no duplicate in the arrays, and they are not sorted and the order of the values must be kept in the main array after modification. I know that can be achieved using such np.setdiff1d or np.in1d (which both are not supported for numba jitted in no-python mode), and other similar posts (e.g. this) have not much more efficient way to do so, but performance is important here because all the values in the main index array will be gradually be removed in loops.
import numpy as np
import numba as nb
n = 500000
r = 10
arr1 = np.random.permutation(n)
arr2 = np.random.randint(0, n, r)
# #nb.jit
def setdif1d_np(a, b):
return np.setdiff1d(a, b, assume_unique=True)
# #nb.jit
def setdif1d_in1d_np(a, b):
return a[~np.in1d(a, b)]
There is another related post that proposed by norok2 for 2D arrays, that is ~15 times faster solution (hashing-like way using numba) than usual methods described there. This solution may be the best if it could be prepared for 1D arrays:
#nb.njit
def mul_xor_hash(arr, init=65537, k=37):
result = init
for x in arr.view(np.uint64):
result = (result * k) ^ x
return result
#nb.njit
def setdiff2d_nb(arr1, arr2):
# : build `delta` set using hashes
delta = {mul_xor_hash(arr2[0])}
for i in range(1, arr2.shape[0]):
delta.add(mul_xor_hash(arr2[i]))
# : compute the size of the result
n = 0
for i in range(arr1.shape[0]):
if mul_xor_hash(arr1[i]) not in delta:
n += 1
# : build the result
result = np.empty((n, arr1.shape[-1]), dtype=arr1.dtype)
j = 0
for i in range(arr1.shape[0]):
if mul_xor_hash(arr1[i]) not in delta:
result[j] = arr1[i]
j += 1
return result
I tried to prepare that for 1D arrays, but I have some problems/question with that.
At first, IDU what does mul_xor_hash exactly do, and if init and k are arbitrary selected or not
Why mul_xor_hash will not work without nb.njit:
File "C:/Users/Ali/Desktop/test - Copy - Copy.py", line 21, in mul_xor_hash
result = (result * k) ^ x
TypeError: ufunc 'bitwise_xor' not supported for the input types, and the inputs could not be safely coerced to any supported types according to the casting rule ''safe''
IDK how to implement mul_xor_hash on 1D arrays (if it could), which I guess may make it faster more than for 2Ds, so I broadcast the input arrays to 2D by [None, :], which get the following error just for arr2:
print(mul_xor_hash(arr2[0]))
ValueError: new type not compatible with array
and what does delta do
I am searching the most efficient way in this regard. In the absence of better method than norok2 solution, how to prepare this solution for 1D arrays?

Understanding the hash-based solution
At first, IDU what does mul_xor_hash exactly do, and if init and k are arbitrary selected or not
mul_xor_hash is a custom hash function. Functions mixing xor and multiply (possibly with shifts) are known to be relatively fast to compute the hash of a raw data buffer. The multiplication tends to shuffle bits and the xor is used to somehow combine/accumulate the result in a fixed size small value (ie. the final hash). There are many different hashing functions. Some are faster than others, some cause more collisions than other in a given context. A fast hashing function causing too many collisions can be useless in practice as it would result in a pathological situation where all conflicting values needs to be compared. This is why fast hash functions are hard to implement.
init and k are parameter certainly causing the hash to be pretty balance. This is pretty common in such a hash function. k needs to be sufficiently big for the multiplication to shuffle bits and it should typically also be a prime number (values like power of two tends to increase collisions due to modular arithmetic behaviours). init plays a significant role only for very small arrays (eg. with 1 item): it helps to reduce collisions by xoring the final hash by a non-trivial constant. Indeed, if arr.size = 1, then result = (init * k) ^ arr[0] where init * k is a constant. Having an identity hash function equal to arr[0] is known to be bad since it tends to result in many collisions (this is a complex topic, but put it shortly, arr[0] can be divided by the number of buckets in the hash table for example). Thus, init should be a relatively big number and init * k should also be a big non-trivial value (a prime number is a good target value).
Why mul_xor_hash will not work without nb.njit
It depends of the input. The input needs to be a 1D array and have a raw size in byte divisible by 8 (eg. 64-bit items, 2n x 32-bit ones, 4n x 16-bit one or 8n 8-bit ones). Here is some examples:
mul_xor_hash(np.random.rand(10))
mul_xor_hash(np.arange(10)) # Do not work with 9
and what does delta do
It is a set containing the hash of the arr2 row so to find matching lines faster than comparing them without hashes.
how to prepare this solution for 1D arrays?
AFAIK, hashes are only use to avoid comparisons of rows but this is because the input is the 2D array. In 1D, there is no such a problem.
There is big catch with this method: it only works if there is no hash collisions. Otherwise, the implementation wrongly assumes that values are equal even if they are not! #norok explicitly mentioned it in the comments though:
Note that the collision handling for the hashings should also be implemented
Faster implementation
Using the 2D solution of #norok2 for 1D is not a good idea since hashes will not make it faster the way they are used. In fact, a set already use a hash function internally anyway. Not to mention collisions needs to be properly implemented (which is done by a set).
Using a set is a relatively good idea since it causes the complexity to be O(n + m) where n = len(arr1) and m = len(arr2). That being said, if arr1 is converted to a set, then it will be too big to fit in L1 cache (due to the size of arr1 in your case) resulting in slow cache misses. Additionally, the growing size of the set will cause values to be re-hashed which is not efficient. If arr2 is converted to a set, then the many hash table fetches will not be very efficient since arr2 is very small in your case. This is why this solution is sub-optimal.
One solution is to split arr1 in chunks and then build a set based on the target chunk. You can then check if a value is in the set or not efficiently. Building the set is still not very efficient due to the growing size. This problem is due to Python itself which do not provide a way to reserve some space for the data structure like other languages do (eg. C++). One solution to avoid this issue is simply to reimplement an hash-table which is not trivial and cumbersome. Actually, Bloom filters can be used to speed up this process since they can quickly find if there is no collision between the two sets arr1 and arr2 in average (though they are not trivial to implement).
Another optimization is to use multiple threads to compute the chunks in parallel since they are independent. That being said, the appending to the final array is not easy to do efficiently in parallel, especially since you do not want the order to be modified. One solution is to move away the copy from the parallel loop and do it serially but this is slow and AFAIK there is no simple way to do that in Numba currently (since the parallelism layer is very limited). Consider using native languages like C/C++ for an efficient parallel implementation.
In the end, hashing can be pretty complex and the speed up can be quite small compared to a naive implementation with two nested loops since arr2 only have few items and modern processors can compare values quickly using SIMD instructions (while hash-based method can hardly benefit from them on mainstream processors). Unrolling can help to write a pretty simple and fast implementation. Again, unfortunately, Numba use LLVM-Jit internally which appear to fail to vectorize such a simple code (certainly due to missing optimizations in either LLVM-Jit or even LLVM itself). As a result, the non vectorized code is finally a bit slower (rather than 4~10 times faster on a modern mainstream processor). One solution is to use a C/C++ code instead to do that (or possibly Cython).
Here is a serial implementation using basic Bloom filters:
#nb.njit('uint32(int32)')
def hash_32bit_4k(value):
return (np.uint32(value) * np.uint32(27_644_437)) & np.uint32(0x0FFF)
#nb.njit(['int32[:](int32[:], int32[:])', 'int32[:](int32[::1], int32[::1])'])
def setdiff1d_nb_faster(arr1, arr2):
out = np.empty_like(arr1)
bloomFilter = np.zeros(4096, dtype=np.uint8)
for j in range(arr2.size):
bloomFilter[hash_32bit_4k(arr2[j])] = True
cur = 0
for i in range(arr1.size):
# If the bloom-filter value is true, we know arr1[i] is not in arr2.
# Otherwise, there is maybe a false positive (conflict) and we need to check to be sure.
if bloomFilter[hash_32bit_4k(arr1[i])] and arr1[i] in arr2:
continue
out[cur] = arr1[i]
cur += 1
return out[:cur]
Here is an untested variant that should work for 64-bit integers (floating point numbers need memory views and possibly a prime constant too):
#nb.njit('uint64(int64)')
def hash_32bit_4k(value):
return (np.uint64(value) * np.uint64(67_280_421_310_721)) & np.uint64(0x0FFF)
Note that if all the values in the small array are contained in the main array in each loop, then we can speed up the arr1[i] in arr2 part by removing values from arr2 when we find them. That being said, collisions and findings should be very rare so I do not expect this to be significantly faster (not to mention it adds some overhead and complexity). If items are computed in chunks, then the last chunks can be directly copied without any check but the benefit should still be relatively small. Note that this strategy can be effective for the naive (C/C++) SIMD implementation previously mentioned though (it can be about 2x faster).
Generalization and parallel implementation
This section focus on the algorithm to use regarding the input size. It particularly details an SIMD-based implementation and discuss about the use of multiple threads.
First of all, regarding the value r, the best algorithm to use can be different. More specifically:
when r is 0, the best thing to do is to return the input array arr1 unmodified (possibly a copy to avoid issue with in-place algorithms);
when r is 1, we can use one basic loop iterating over the array, but the best implementation is likely to use np.where of Numpy which is highly optimized for that
when r is small like <10, then using a SIMD-based implementation should be particularly efficient, especially if the iteration range of the arr2-based loop is known at compile-time and is unrolled
for bigger r values that are still relatively small (eg. r < 1000 and r << n), the provided hash-based solution should be one of the best;
for larger r values with r << n, the hash-based solution can be optimized by packing boolean values as bits in bloomFilter and by using multiple hash-functions instead of one so to better handle collisions while being more cache-friendly (in fact, this is what actual bloom filters does); note that multi-threading can be used so speed up the lookups when r is huge and r << n;
when r is big and not much smaller than n, then the problem is pretty hard to solve efficiently and the best solution is certainly to sort both arrays (typically with a radix sort) and use a merge-based method to remove the duplicates, possibly with multiple threads when both r and n are huge (hard to implement).
Let's start with the SIMD-based solution. Here is an implementation:
#nb.njit('int32[:](int32[::1], int32[::1])')
def setdiff1d_nb_simd(arr1, arr2):
out = np.empty_like(arr1)
limit = arr1.size // 4 * 4
limit2 = arr2.size // 2 * 2
cur = 0
z32 = np.int32(0)
# Tile (x4) based computation
for i in range(0, limit, 4):
f0, f1, f2, f3 = z32, z32, z32, z32
v0, v1, v2, v3 = arr1[i], arr1[i+1], arr1[i+2], arr1[i+3]
# Unrolled (x2) loop searching for a match in `arr2`
for j in range(0, limit2, 2):
val1 = arr2[j]
val2 = arr2[j+1]
f0 += (v0 == val1) + (v0 == val2)
f1 += (v1 == val1) + (v1 == val2)
f2 += (v2 == val1) + (v2 == val2)
f3 += (v3 == val1) + (v3 == val2)
# Remainder of the previous loop
if limit2 != arr2.size:
val = arr2[arr2.size-1]
f0 += v0 == val
f1 += v1 == val
f2 += v2 == val
f3 += v3 == val
if f0 == 0: out[cur] = arr1[i+0]; cur += 1
if f1 == 0: out[cur] = arr1[i+1]; cur += 1
if f2 == 0: out[cur] = arr1[i+2]; cur += 1
if f3 == 0: out[cur] = arr1[i+3]; cur += 1
# Remainder
for i in range(limit, arr1.size):
if arr1[i] not in arr2:
out[cur] = arr1[i]
cur += 1
return out[:cur]
It turns out this implementation is always slower than the hash-based one on my machine since Numba clearly generate an inefficient for the inner arr2-based loop and this appears to come from broken optimizations related to the ==: Numba simply fail use SIMD instructions for this operation (for no apparent reasons). This prevent many alternative SIMD-related codes to be fast as long as they are using Numba.
Another issue with Numba is that np.where is slow since it use a naive implementation while the one of Numpy has been heavily optimized. The optimization done in Numpy can hardly be applied to the Numba implementation due to the previous issue. This prevent any speed up using np.where in a Numba code.
In practice, the hash-based implementation is pretty fast and the copy takes a significant time on my machine already. The computing part can be speed up using multiple thread. This is not easy since the parallelism model of Numba is very limited. The copy cannot be easily optimized with Numba (one can use non-temporal store but this is not yet supported by Numba) unless the computation is possibly done in-place.
To use multiple threads, one strategy is to first split the range in chunk and then:
build a boolean array determining, for each item of arr1, whether the item is found in arr2 or not (fully parallel)
count the number of item found by chunk (fully parallel)
compute the offset of the destination chunk (hard to parallelize, especially with Numba, but fast thanks to chunks)
copy the chunk to the target location without copying found items (fully parallel)
Here is an efficient parallel hash-based implementation:
#nb.njit('int32[:](int32[:], int32[:])', parallel=True)
def setdiff1d_nb_faster_par(arr1, arr2):
# Pre-computation of the bloom-filter
bloomFilter = np.zeros(4096, dtype=np.uint8)
for j in range(arr2.size):
bloomFilter[hash_32bit_4k(arr2[j])] = True
chunkSize = 1024 # To tune regarding the kind of input
chunkCount = (arr1.size + chunkSize - 1) // chunkSize
# Find for each item of `arr1` if the value is in `arr2` (parallel)
# and count the number of item found for each chunk on the fly.
# Note: thanks to page fault, big parts of `found` are not even written in memory if `arr2` is small
found = np.zeros(arr1.size, dtype=nb.bool_)
foundCountByChunk = np.empty(chunkCount, dtype=nb.uint16)
for i in nb.prange(chunkCount):
start, end = i * chunkSize, min((i + 1) * chunkSize, arr1.size)
foundCountInChunk = 0
for j in range(start, end):
val = arr1[j]
if bloomFilter[hash_32bit_4k(val)] and val in arr2:
found[j] = True
foundCountInChunk += 1
foundCountByChunk[i] = foundCountInChunk
# Compute the location of the destination chunks (sequential)
outChunkOffsets = np.empty(chunkCount, dtype=nb.uint32)
foundCount = 0
for i in range(chunkCount):
outChunkOffsets[i] = i * chunkSize - foundCount
foundCount += foundCountByChunk[i]
# Parallel chunk-based copy
out = np.empty(arr1.size-foundCount, dtype=arr1.dtype)
for i in nb.prange(chunkCount):
srcStart, srcEnd = i * chunkSize, min((i + 1) * chunkSize, arr1.size)
cur = outChunkOffsets[i]
# Optimization: we can copy the whole chunk if there is nothing found in it
if foundCountByChunk[i] == 0:
out[cur:cur+(srcEnd-srcStart)] = arr1[srcStart:srcEnd]
else:
for j in range(srcStart, srcEnd):
if not found[j]:
out[cur] = arr1[j]
cur += 1
return out
This implementation is the fastest for the target input on my machine. It is generally fast when n is quite big and the overhead to create threads is relatively small on the target platform (eg. on PCs but typically not computing servers with many cores). The overhead of the parallel implementation is significant so the number of core on the target machine needs to be at least 4 so the implementation can be significantly faster than the sequential implementation.
It may be useful to tune the chunkSize variable for the target inputs. If r << n, it is better to use a pretty big chunkSize. That being said, the number of chunk needs to be sufficiently big for multiple thread to operate on many chunks. Thus, chunkSize should be significantly smaller than n / numberOfThreads.
On my machine most of the time (65-70%) is spent in the final copy which is mostly memory-bound and can hardly be optimized further with Numba.
Results
Here are results on my i5-9600KF-based machine (with 6 cores):
setdif1d_np: 2.65 ms
setdif1d_in1d_np: 2.61 ms
setdiff1d_nb: 2.33 ms
setdiff1d_nb_simd: 1.85 ms
setdiff1d_nb_faster: 0.73 ms
setdiff1d_nb_faster_par: 0.49 ms
The best provided implementation is about 4~5 time faster than the other ones.

What I found is that hashing does not help,. It is just trick for 2D case, to convert 1d arrays to single numbers and put them as such in a set.
Below is method of norok2 I converted to 1d arrays (and added annotations for faster compilation).
Note that this is only slightly (20-30%) faster than the methods you already have. And of course after second function call, on first due to compilation it is slightly slower.
#nb.njit('int32[:](int32[:], int32[:])')
def setdiff1d_nb(arr1, arr2):
delta = set(arr2)
# : build the result
result = np.empty(len(arr1), dtype=arr1.dtype)
j = 0
for i in range(arr1.shape[0]):
if arr1[i] not in delta:
result[j] = arr1[i]
j += 1
return result[:j]

Is MATLAB's bsxfun the best? Python's numpy.einsum?

I have a very large multiply and sum operation that I need to implement as efficiently as possible. The best method I've found so far is bsxfun in MATLAB, where I formulate the problem as:
L = 10000;
x = rand(4,1,L+1);
A_k = rand(4,4,L);
tic
for k = 2:L
i = 2:k;
x(:,1,k+1) = x(:,1,k+1)+sum(sum(bsxfun(#times,A_k(:,:,2:k),x(:,1,k+1-i)),2),3);
end
toc
Note that L will be larger in practice. Is there a faster method? It's strange that I need to first add the singleton dimension to x and then sum over it, but I can't get it to work otherwise.
It's still much faster than any other method I've tried, but not enough for our application. I've heard rumors that the Python function numpy.einsum may be more efficient, but I wanted to ask here first before I consider porting my code.
I'm using MATLAB R2017b.

I believe both of your summations can be removed, but I only removed the easier one for the time being. The summation over the second dimension is trivial, since it only affects the A_k array:
B_k = sum(A_k,2);
for k = 2:L
i = 2:k;
x(:,1,k+1) = x(:,1,k+1) + sum(bsxfun(#times,B_k(:,1,2:k),x(:,1,k+1-i)),3);
end
With this single change the runtime is reduced from ~8 seconds to ~2.5 seconds on my laptop.
The second summation could also be removed, by transforming times+sum into a matrix-vector product. It needs some singleton fiddling to get the dimensions right, but if you define an auxiliary array that is B_k with the second dimension reversed, you can generate the remaining sum as ~x*C_k with this auxiliary array C_k, give or take a few calls to reshape.
So after a closer look I realized that my original assessment was overly optimistic: you have multiplications in both dimensions in your remaining term, so it's not a simple matrix product. Anyway, we can rewrite that term to be the diagonal of a matrix product. This implies that we're computing a bunch of unnecessary matrix elements, but this still seems to be slightly faster than the bsxfun approach, and we can get rid of your pesky singleton dimension too:
L = 10000;
x = rand(4,L+1);
A_k = rand(4,4,L);
B_k = squeeze(sum(A_k,2)).';
tic
for k = 2:L
ii = 1:k-1;
x(:,k+1) = x(:,k+1) + diag(x(:,ii)*B_k(k+1-ii,:));
end
toc
This runs in ~2.2 seconds on my laptop, somewhat faster than the ~2.5 seconds obtained previously.

Since you're using an new version of Matlab you might try broadcasting / implicit expansion instead of bsxfun:
x(:,1,k+1) = x(:,1,k+1)+sum(sum(A_k(:,:,2:k).*x(:,1,k-1:-1:1),3),2);
I also changed the order of summation and removed the i variable for further improvement. On my machine, and with Matlab R2017b, this was about 25% faster for L = 10000.

List comprehension for np matrixes

I have two np.matrixes, one of which I'm trying to normalize. I know, in general, list comprehensions are faster than for loops, so I'm trying to convert my double for loop into a list expression.
# normalize the rows and columns of A by B
for i in range(1,q+1):
for j in range(1,q+1):
A[i-1,j-1] = A[i-1,j-1] / (B[i-1] / B[j-1])
This is what I have gotten so far:
A = np.asarray([A/(B[i-1]/B[j-1]) for i, j in zip(range(1,q+1), range(1,q+1))])
but I think I'm taking the wrong approach because I'm not seeing any significant time difference.
Any help would be appreciated.

First, if you really do mean np.matrix, stop using np.matrix. It has all sorts of nasty incompatibilities, and its role is obsolete now that # for matrix multiplication exists. Even if you're stuck on a Python version without #, using the dot method with normal ndarrays is still better than dealing with np.matrix.
You shouldn't use any sort of Python-level iteration construct with NumPy arrays, whether for loops or list comprehensions, unless you're sure you have no better options. Assuming A is 2D and B is 1D with shapes (q, q) and (q,) respectively, what you should instead do for this case is
A *= B
A /= B[:, np.newaxis]
broadcasting the operation over A. This will allow NumPy to perform the iteration at C level directly over the arrays' underlying data buffers, without having to create wrapper objects and perform dynamic dispatch on every operation.

I don't understand why/how one of these methods is faster than the others

I wanted to test the difference in time between implementations of some simple code. I decided to count how many values out of a random sample of 10,000,000 numbers is greater than 0.5. The random sample is grabbed uniformly from the range [0.0, 1.0).
Here is my code:
from numpy.random import random_sample; import time;
n = 10000000;
t1 = time.clock();
t = 0;
z = random_sample(n);
for x in z:
if x > 0.5: t += 1;
print t;
t2 = time.clock();
t = 0;
for _ in xrange(n):
if random_sample() > 0.5: t += 1;
print t;
t3 = time.clock();
t = (random_sample(n) > 0.5).sum();
print t;
t4 = time.clock();
print t2-t1; print t3-t2; print t4-t3;
This is the output:
4999445
4999511
5001498
7.0348236652
1.75569394301
0.202538106332
I get that the first implementation sucks because creating a massive array and then counting it element-wise is a bad idea, so I thought that the second implementation would be the most efficient.
But how is the third implementation 10 times faster than the second method? Doesn't the third method also create a massive array in the form of random_sample(n) and then go through it checking values against 0.5?
How is this third method different from the first method and why is it ~35 times faster than the first method?
EDIT: #merlin2011 suggested that Method 3 probably doesn't create the full array in memory. So, to test that theory I tried the following:
z = random_sample(n);
t = (z > 0.5).sum();
print t;
which runs in a time of 0.197948451549 which is practically identical to Method 3. So, this is probably not a factor.

Method 1 generates a full list in memory before using it. This is slow because the memory has to be allocated and then accessed, probably missing the cache multiple times.
Method 2 uses an generator, which never creates the list in memory but instead generates each element on demand.
Method 3 is probably faster because sum() is implemented as a loop in C but I am not 100% sure. My guess is that this is faster for the same reason that Matlab vectorization is faster than for loops in Matlab.
Update: Separating out each of three steps, I observe that method 3 is still equally fast, so I have to agree with utdemir that each individual operator is executing instructions closer to machine code.
z = random_sample(n)
z2 = z > 0.5
t = z2.sum();
In each of the first two methods, you are invoking Python's standard functionality to do a loop, and this is much slower than a C-level loop that is baked into the implementation.

AFAIK
Function calls are heavy, on method two, you're calling random_sample() 10000000 times, but on third method, you just call it once.
Numpy's > and .sum are optimized to their last bits in C, also most probably using SIMD instructions to avoid loops.
So,
On method 2, you are comparing and looping using Python; but on method 3, you're much closer to the processor and using optimized instructions to compare and sum.